MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $A$ a (commutative) ring, $f:A\to B$ an $A$-algebra, what conditions do we need on $A$ and $B$ (and $f$) for the functor $-\otimes_A B:A-mod\to B-mod\quad$ to be faithful (i.e. injective on $Hom$-sets)?

I can't seem to come up with anything other than the rather obvious condition that tensoring with $B$ shouldn't kill anything (or at least not too much), but this is hardly satisfying. It seems like it would be faithful in general, but I fail to come up with an argument as to why this should be true, and precisely when it is (if at all). Dare I beg the aid of the MO?

Note: this is basically the same as $f^*$ being faithful for a morphism $f:X\to Y$ of schemes, which reduces to the above. Hence the algebraic geometry tag.

share|cite|improve this question
Certainly not faithful in general. Lot of examples when $A$ is local, but $f$ is not a local homomorphism. For example, it could be the inclusion of a domain into its quotient field. Then all the torsion stuff will die, definitely non-faithful. – Graham Leuschke Dec 2 '10 at 14:22
up vote 11 down vote accepted

The functor you mention is faithful if and only if the functor $-\bigotimes_A B :A-mod\to A-mod$ is faithful, ie iff $B$ is a faithful $A$-module.

For a concrete counterexample take $f:\mathbb{Z}\to \mathbb{Q}$ and like Graham says this kills the torsion stuff.

share|cite|improve this answer
Why must $B$ be flat? Seems like this is the content of the earlier question -- is it possible for $-\otimes_AB$ to be faithful without $B$ being flat. – Graham Leuschke Dec 2 '10 at 15:55
Yes, it is possible to be faithful without being flat. Let $A=k[x,y]$ and let $B= xA+yA$. – Greg Muller Dec 2 '10 at 17:29
Yes. I have now replaced "faithfully flat" to "faithful" in my answer. – Achilleas K Dec 2 '10 at 17:41
@Greg, do you know of any examples of commutative rings $B$ with unity? – Karl Schwede Dec 2 '10 at 17:57
I had missed the requirement that $B$ was a ring, rather than an $A$-module. Of course, I can replace $B$ by its symmetric tensor algebra to get a ring which is faithful but not fat. – Greg Muller Dec 2 '10 at 18:44

A morphism $f:X\rightarrow Y$ of schemes gives a faithful pullback functor $f^*$ exactly when the morphism $f$ is surjective on underlying sets. This can be seen in several steps.

  1. Note that the functor is faithful iff it preserves zero; that is, $f^*(M)=0$ implies $M=0$.
    a.To see this, note that if a right exact functor preserves zero, then any morphism which becomes the zero map was already the zero map (consider its cokernel). b. Then, if two maps $g,g'$ go to the same map, their difference goes to the zero map, and thus the difference was already zero.

  2. A module $M$ on $Y$ is zero iff $Hom_Y(\\mathcal{O}_Z,M)=0$ for every irreducible closed subscheme $Z$ (that is, for the closure of every point).

  3. If $Hom_Y(\mathcal{O}_Z,f^*(M))=0$, then $Hom_X(\mathcal{O}_{f(Z)},M)=0$.

Now, if every point in $X$ is the image of a point in $Y$, and $f^*(M)=0$, then by 2 and 3, $Hom_X(\mathcal{O}_{Z'},M)=0$ for every point $Z'\in X$. Then, by $2$, we know that $M=0$, and so $f^* $ preserves zero. Thus, by 1, $f^* $ is faithful. If $f$ is not surjective on underlying sets, then pulling back the skyscraper sheaf of a missed point will be zero, and so by 1, it cannot be faithful.

share|cite|improve this answer
I'd give this an extra +1 if I could, as it also answers the problem I had which motivated the thread. Unfortunately, the answer was negative. – Ketil Tveiten Dec 2 '10 at 18:19
I don't believe that 2. is true. Since $\mathcal{O}_Z$ is a quotient of $\mathcal{O}_Y$, already $\Gamma(X,M) = 0$ implies $\mathrm{Hom}_X(\mathcal{O}_Z,M)=0$. Or do you mean the Hom-sheaf? – Martin Brandenburg Nov 12 '11 at 12:13
Also the last step of the proof seems to be problematic if the point is not closed: How do you prove that the skyscraper sheaf pulls back to $0$? – Martin Brandenburg Nov 12 '11 at 12:40
Here is a counterexample to the statement "f surjective implies $f^*$ faithful": It is enough to find a nil ideal $I$ of some ring $A$ and an $A$-module $M$ such that $M = IM$, but $M \neq 0$; then $\mathrm{Spec}(A/I) \to \mathrm{Spec}(A)$ is a homeomorphism, but the corresponding pullback functor is not faithful. One specific example is $A = k[x_1,x_2,\dotsc]/(x_i^i = 0, x_i = x_{i+1} x_{i+2})_{i \geq 1}, M = I = (x_1,x_2,\dotsc)$. – Martin Brandenburg Nov 12 '11 at 15:44
A more easy counterexmaple: The zero section $\mathrm{Spec}(k) \to \mathrm{Spec}(k[e]/e^2)$. – Martin Brandenburg Nov 15 '11 at 8:28

A sufficient condition is that the embedding $A \to B$ splits in the category of $A$-modules, i.e. $B \cong A \oplus B'$ as $A$-module. In this case the functor $-\otimes_A B$ has $\id$ as a direct summand, hence faithful. An example $A = {\mathbb Z}$, $B = {\mathbb Z} \oplus {\mathbb Z}/2{\mathbb Z}$ shows that one does not need to require that $B$ is flat.

On the other hand, if you want $-\otimes_A B$ to be faithful also on $Ext$'s, then the above condition is also necessary. Indeed, consider exact sequence $$ 0 \to A \to B \to B/A \to 0. $$ It gives an extension of $B/A$ by $A$. If we tensor it with $B$ we obtain $$ B \to B\otimes_A B \to (B/A)\otimes_A B \to 0. $$ Note that the first map is a split monomorphism (because of the multiplication $B\otimes_A B \to B$), hence the corresponding extension is trivial. So, if we want the functor to be injective on extensions the initial exact sequence should be split, hence $B = A \oplus (B/A)$.

share|cite|improve this answer
What do you mean exactly by "injective on Exts" or "injective on extensions"? – Martin Brandenburg Nov 12 '11 at 15:56

The answer is given in the following - very interesting - paper:

Bachuki Mesablishvili, Descent Theory for Schemes, Applied Categorical Structures 12: 485–512, 2004.

The author generalizes Grothendieck's descent theory from faithfully flat morphisms to pure morphisms. These are morphisms of schemes $f$ such that every base change $f'$ of it is schematically dense in the sense that $f'$ does not factor through a proper subscheme. In the affine case, $\mathrm{Spec}(A) \to \mathrm{Spec}(R)$ is pure iff $R \to A$ is "stable injective" in the sense that for every $R$-algebra $B$ the map $B \to A \otimes_R B$ is injective.

One of the main results (Theorem 5.15) states that for a quasi-compact morphism $f : X \to Y$ the following are equivalent:

1) $f$ is pure

2) $f^\* : \mathrm{Qcoh}(Y) \to \mathrm{Qcoh}(X)$ faithful

3) $f$ is a stable effective descent morphism for quasi-coherent sheaves

In the case that $X,Y$ are affine there are even more characterizations (Theorem 4.18), for example we can also add 4) $f^\*$ is conservative, and 5) $f^\*$ is comonadic.

For flat morphisms, it is well-known that $f^*$ is faithful iff $f$ is surjective iff $f$ is faithfully flat. In general, every pure morphism is surjective (since otherwise some base change $\emptyset \to \mathrm{Spec}(\text{field})$ is not schematically dense). But there are lots of surjective morphisms which are not pure, for example the zero section $s: \mathrm{Spec}(k) \to \mathrm{Spec}(k[\epsilon]/\epsilon^2)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.